Relevance for Computer Algebra We desire algorithms ◮ proving ◮ finding ◮ using algebraic relations for specific classes of sequences. Applications: Summation/Integration of special functions. We have an algorithm that can find identities like n √ √ √ � (( k − k + 1) H k + 1) k ! = (1 + ( n + 1) H n ) n ! k =0 which depend on exploiting the relation √ n 2 − n = 0 .
Relevance for Computer Algebra We desire algorithms ◮ proving ◮ finding ◮ using algebraic relations for specific classes of sequences. Applications: Summation/Integration of special functions. We want to have an algorithm that can find identities like n (( − 1) k − 1) x +( − 1) k +1 = (1 − 2 x ) U n ( x )+( − 1) n + U n +1 ( x ) � 2 U k ( x )+( − 1) k − 1 2 U n ( x )+( − 1) n − 1 k =0 which also depend on nontrivial relations.
Relevance for Computer Algebra We desire algorithms ◮ proving ◮ finding ◮ using algebraic relations for specific classes of sequences. Today we discuss finding.
Problem Specification GIVEN: Sequences f 1 ( n ) , . . . , f m ( n ) in �
Problem Specification GIVEN: Sequences f 1 ( n ) , . . . , f m ( n ) in � � [ x 1 , . . . , x m ] such that FIND: Polynomials b 1 , . . . , b r ∈ � [ x 1 , . . . , x m ] � b 1 , . . . , b r � � is the ideal of algebraic relations among the f i ( n ) .
Problem Specification GIVEN: Sequences f 1 ( n ) , . . . , f m ( n ) in � � [ x 1 , . . . , x m ] such that FIND: Polynomials b 1 , . . . , b r ∈ � [ x 1 , . . . , x m ] � b 1 , . . . , b r � � is the ideal of algebraic relations among the f i ( n ) . Of course, “given sequences” makes only sense when attention is restricted to particular classes of sequences that admit finitary representations, e.g., by defining recurrence equations.
A Brute Force Attack
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Consider the polynomial p ( x 1 , x 2 ) = a 0 x 2 1 + a 1 x 1 x 2 + a 2 x 2 2 + a 3 x 1 + a 4 x 2 + a 5 with undetermined coefficients a 0 , a 1 , . . . , a 5 .
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. For p ( x 1 , x 2 ) to be an algebraic relation of f 1 ( n ) , f 2 ( n ) , we must have ∀ n ≥ 0 : p ( f 1 ( n ) , f 2 ( n )) = 0
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. In particular: p ( f 1 (0) , f 2 (0)) = 0 p ( f 1 (1) , f 2 (1)) = 0 p ( f 1 (2) , f 2 (2)) = 0 . . . p ( f 1 (5) , f 2 (5)) = 0
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. In particular: a 0 f 1 (0) 2 + a 1 f 1 (0) f 2 (0) + a 2 f 2 (0) 2 + a 3 f 1 (0) + a 4 f 2 (0) + a 5 = 0 a 0 f 1 (1) 2 + a 1 f 1 (1) f 2 (1) + a 2 f 2 (1) 2 + a 3 f 1 (1) + a 4 f 2 (1) + a 5 = 0 a 0 f 1 (2) 2 + a 1 f 1 (2) f 2 (2) + a 2 f 2 (2) 2 + a 3 f 1 (2) + a 4 f 2 (2) + a 5 = 0 . . . a 0 f 1 (5) 2 + a 1 f 1 (5) f 2 (5) + a 2 f 2 (5) 2 + a 3 f 1 (5) + a 4 f 2 (5) + a 5 = 0
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Solutions of the linear system give relation candidates.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Solutions of the linear system give relation candidates. True relations can be detected by the prover.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Solutions of the linear system give relation candidates. True relations can be detected by the prover. If there are fake ones, repeat with more sample points.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Solutions of the linear system give relation candidates. True relations can be detected by the prover. If there are fake ones, repeat with more sample points. If sufficiently many points are taken into account, no fake relations will arise.
A Brute Force Attack Suppose we know how to prove algebraic relations for a certain class of sequences. Then we can also find algebraic relations. By just making an ansatz. Solutions of the linear system give relation candidates. True relations can be detected by the prover. If there are fake ones, repeat with more sample points. If sufficiently many points are taken into account, no fake relations will arise. For speed-up, use the Buchberger-M¨ oller algorithm.
Example: Fibonacci Numbers Let F n denote the n th Fibonacci number ( n ∈ � ). Exercise 6.81: (Graham/Knuth/Patashnik) Let P ( x, y ) be a polynomial in x and y with integer coefficients. Find a necessary and sufficient condition that P ( F n +1 , F n ) = 0 for all n ≥ 0 . � [ x, y ] of algebraic relations In other words: Find the ideal a � among ( F n ) n ≥ 0 and ( F n +1 ) n ≥ 0 .
Example: Fibonacci Numbers Let F n denote the n th Fibonacci number ( n ∈ � ). Exercise 6.81: (Graham/Knuth/Patashnik) Let P ( x, y ) be a polynomial in x and y with integer coefficients. Find a necessary and sufficient condition that P ( F n +1 , F n ) = 0 for all n ≥ 0 . � [ x, y ] of algebraic relations In other words: Find the ideal a � among ( F n ) n ≥ 0 and ( F n +1 ) n ≥ 0 .
Example: Fibonacci Numbers Let F n denote the n th Fibonacci number ( n ∈ � ). Exercise 6.81: (Graham/Knuth/Patashnik) Let P ( x, y ) be a polynomial in x and y with integer coefficients. Find a necessary and sufficient condition that P ( F n +1 , F n ) = 0 for all n ≥ 0 . � [ x, y ] of algebraic relations In other words: Find the ideal a � among ( F n ) n ≥ 0 and ( F n +1 ) n ≥ 0 . Answer: a ⊇ � ( x 2 − xy − y 2 − 1)( x 2 − xy − y 2 + 1) � .
Example: Fibonacci Numbers Let F n denote the n th Fibonacci number ( n ∈ � ). Exercise 6.81: (Graham/Knuth/Patashnik) Let P ( x, y ) be a polynomial in x and y with integer coefficients. Find a necessary and sufficient condition that P ( F n +1 , F n ) = 0 for all n ≥ 0 . � [ x, y ] of algebraic relations In other words: Find the ideal a � among ( F n ) n ≥ 0 and ( F n +1 ) n ≥ 0 . Answer: a ⊇ � ( x 2 − xy − y 2 − 1)( x 2 − xy − y 2 + 1) � . Note: We can determine all algebraic relations up to a prescribed degree, but we get no information about existence/non-existence of higher degree relations.
Example: Somos Sequences A sequence C n satisfying a nonlinear recurrence of the form C n + r C n = α 1 C n + r − 1 C n +1 + α 2 C n + r − 2 C n +2 + · · · · · · + α ⌊ r/ 2 ⌋ C n + r −⌊ r/ 2 ⌋ C n + ⌊ r/ 2 ⌋ � fixed and α 1 , . . . , α ⌊ r/ 2 ⌋ is called a Somos sequence of with r ∈ order r .
Example: Somos Sequences A sequence C n satisfying a nonlinear recurrence of the form C n + r C n = α 1 C n + r − 1 C n +1 + α 2 C n + r − 2 C n +2 + · · · · · · + α ⌊ r/ 2 ⌋ C n + r −⌊ r/ 2 ⌋ C n + ⌊ r/ 2 ⌋ � fixed and α 1 , . . . , α ⌊ r/ 2 ⌋ is called a Somos sequence of with r ∈ order r . Question: Can a given Somos sequence of order r also be viewed as a Somos sequence for some different order r ′ ?
Example: Somos Sequences A sequence C n satisfying a nonlinear recurrence of the form C n + r C n = α 1 C n + r − 1 C n +1 + α 2 C n + r − 2 C n +2 + · · · · · · + α ⌊ r/ 2 ⌋ C n + r −⌊ r/ 2 ⌋ C n + ⌊ r/ 2 ⌋ � fixed and α 1 , . . . , α ⌊ r/ 2 ⌋ is called a Somos sequence of with r ∈ order r . Question: Can a given Somos sequence of order r also be viewed as a Somos sequence for some different order r ′ ? Example: Consider C n defined via C n +4 C n = C n +3 C n +1 + C 2 n +2 , C 0 = C 1 = C 2 = C 3 = 1 . Does this sequence satisfy a Somos-like recurrence of orders 5 , 6 , 7 , 8 ?
Example: Somos Sequences Idea: Compute the algebraic relations of total degree ≤ 2 among the terms C n , C n +1 , . . . , C n +7 , C n +8 .
Example: Somos Sequences Idea: Compute the algebraic relations of total degree ≤ 2 among the terms C n , C n +1 , . . . , C n +7 , C n +8 . Let a = � p 1 , . . . , p k � � � [ x 0 , . . . , x 8 ] be a Gr¨ obner basis for the ideal generated by the quadratic relations.
Example: Somos Sequences Idea: Compute the algebraic relations of total degree ≤ 2 among the terms C n , C n +1 , . . . , C n +7 , C n +8 . Let a = � p 1 , . . . , p k � � � [ x 0 , . . . , x 8 ] be a Gr¨ obner basis for the ideal generated by the quadratic relations. Make an ansatz with undetermined coefficients for the desired relation, e.g., C n +5 C n = a 1 C n +4 C n +1 + a 2 C n +3 C n +2
Example: Somos Sequences Idea: Compute the algebraic relations of total degree ≤ 2 among the terms C n , C n +1 , . . . , C n +7 , C n +8 . Let a = � p 1 , . . . , p k � � � [ x 0 , . . . , x 8 ] be a Gr¨ obner basis for the ideal generated by the quadratic relations. Make an ansatz with undetermined coefficients for the desired relation, e.g., C n +5 C n = a 1 C n +4 C n +1 + a 2 C n +3 C n +2 Reduction modulo a gives → a (1 − 1 5 a 2 ) x 0 x 5 − ( a 1 + 1 x 5 x 0 − a 1 x 4 x 1 − a 2 x 3 x 2 − 5 a 2 ) x 1 x 4
Example: Somos Sequences Idea: Compute the algebraic relations of total degree ≤ 2 among the terms C n , C n +1 , . . . , C n +7 , C n +8 . Let a = � p 1 , . . . , p k � � � [ x 0 , . . . , x 8 ] be a Gr¨ obner basis for the ideal generated by the quadratic relations. Make an ansatz with undetermined coefficients for the desired relation, e.g., C n +5 C n = a 1 C n +4 C n +1 + a 2 C n +3 C n +2 Reduction modulo a gives → a (1 − 1 5 a 2 ) x 0 x 5 − ( a 1 + 1 x 5 x 0 − a 1 x 4 x 1 − a 2 x 3 x 2 − 5 a 2 ) x 1 x 4 Comparing coefficients gives a 1 = − 1 , a 2 = 5 .
What about a Degree Bound? Note: If the chosen degree bound d is sufficiently large then we get a basis for the whole ideal.
What about a Degree Bound? Note: If the chosen degree bound d is sufficiently large then we get a basis for the whole ideal. But: What does “sufficiently large” mean?
What about a Degree Bound? Note: If the chosen degree bound d is sufficiently large then we get a basis for the whole ideal. But: What does “sufficiently large” mean? In particular: Can we compute a “sufficiently large” d ?
What about a Degree Bound? Note: If the chosen degree bound d is sufficiently large then we get a basis for the whole ideal. But: What does “sufficiently large” mean? In particular: Can we compute a “sufficiently large” d ? well, hardly ever. . .
What about a Degree Bound? Theorem. If a class C of sequences is such that
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C ◮ f ( n ) ∈ C ⇒ � n k =0 f ( k ) ∈ C
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C ◮ f ( n ) ∈ C ⇒ � n k =0 f ( k ) ∈ C ◮ There is an algorithm that produces for arbitrary given f 1 ( n ) , . . . , f m ( n ) ∈ C a basis for their ideal of algebraic relations.
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C ◮ f ( n ) ∈ C ⇒ � n k =0 f ( k ) ∈ C ◮ There is an algorithm that produces for arbitrary given f 1 ( n ) , . . . , f m ( n ) ∈ C a basis for their ideal of algebraic relations. Then there exists an algorithm that decides for arbitrary given f ( n ) ∈ C whether ∃ n ≥ 0 : f ( n ) = 0 .
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C ◮ f ( n ) ∈ C ⇒ � n k =0 f ( k ) ∈ C ◮ There is an algorithm that produces for arbitrary given f 1 ( n ) , . . . , f m ( n ) ∈ C a basis for their ideal of algebraic relations. Then there exists an algorithm that decides for arbitrary given f ( n ) ∈ C whether ∃ n ≥ 0 : f ( n ) = 0 . For sufficiently rich classes C there is no hope for such an algorithm.
What about a Degree Bound? Theorem. If a class C of sequences is such that ◮ n ∈ C ◮ f ( n ) ∈ C ⇒ � n k =0 f ( k ) ∈ C ◮ There is an algorithm that produces for arbitrary given f 1 ( n ) , . . . , f m ( n ) ∈ C a basis for their ideal of algebraic relations. Then there exists an algorithm that decides for arbitrary given f ( n ) ∈ C whether ∃ n ≥ 0 : f ( n ) = 0 . For sufficiently rich classes C there is no hope for such an algorithm. If we insist in a complete algorithm, we have to focus on smaller classes.
C-Finite Sequences (joint work with B. Zimmermann)
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈ Examples:
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈ Examples: ◮ n , n 2 , n 3 , . . .
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈ Examples: ◮ n , n 2 , n 3 , . . . ◮ 2 n , 3 n , 4 n , . . .
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈ Examples: ◮ n , n 2 , n 3 , . . . ◮ 2 n , 3 n , 4 n , . . . ◮ F n , U n ( x ) , . . .
C-finite Sequences Recall: f ( n ) is C-finite if f ( n + r ) = a 0 f ( n ) + a 1 f ( n + 1) + · · · + a r − 1 f ( n + r − 1) � . for some constants a 0 , . . . , a r − 1 ∈ Recall: f ( n ) is C-finite if and only if f ( n ) = p 1 ( n ) φ n 1 + p 2 ( n ) φ n 2 + · · · + p s ( n ) φ n ( n ≥ 0) s where φ i are the roots of the characteristic polynomial x r − a 0 − a 1 x − a 2 x 2 − · · · − a r − 1 x r − 1 and p i ( n ) is a polynomial whose degree is bounded by the multiplicity of the root φ i .
Simple Examples Example 1: n and 2 n are algebraically independent.
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.)
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent.
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .)
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .) Example 3: ( − 1) n is algebraically dependent.
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .) Example 3: ( − 1) n is algebraically dependent. (by x 2 1 − 1 .)
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .) Example 3: ( − 1) n is algebraically dependent. (by x 2 1 − 1 .) Example 4: 4 n , 6 n , 9 n are algebraically dependent.
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .) Example 3: ( − 1) n is algebraically dependent. (by x 2 1 − 1 .) Example 4: 4 n , 6 n , 9 n are algebraically dependent. (by x 1 x 3 − x 2 2 .)
Simple Examples Example 1: n and 2 n are algebraically independent. (clear.) Example 2: n 2 and n 3 are algebraically dependent. (by x 3 1 − x 2 2 .) Example 3: ( − 1) n is algebraically dependent. (by x 2 1 − 1 .) Example 4: 4 n , 6 n , 9 n are algebraically dependent. (by x 1 x 3 − x 2 2 .) Example 5: 4 n , 7 n , 9 n are algebraically independent.
In General: Algebraic Relations of Exponentials Theorem: Let φ 1 , . . . , φ m ∈ �
In General: Algebraic Relations of Exponentials � and Theorem: Let φ 1 , . . . , φ m ∈ L := { ( c 1 , . . . , c m ) : φ c 1 1 φ c 2 2 · · · φ c m � m . m = 1 } ⊆
In General: Algebraic Relations of Exponentials � and Theorem: Let φ 1 , . . . , φ m ∈ L := { ( c 1 , . . . , c m ) : φ c 1 1 φ c 2 2 · · · φ c m � m . m = 1 } ⊆ This is a lattice.
In General: Algebraic Relations of Exponentials � and Theorem: Let φ 1 , . . . , φ m ∈ L := { ( c 1 , . . . , c m ) : φ c 1 1 φ c 2 2 · · · φ c m � m . m = 1 } ⊆ This is a lattice. Let � [ x 1 , . . . , x m ] I ( L ) := � x c 1 1 x c 2 2 · · · x c m m − 1 : ( c 1 , . . . , c m ) ∈ L � � be the corresponding lattice ideal. (It is understood that “denominators are cleared”.)
In General: Algebraic Relations of Exponentials � and Theorem: Let φ 1 , . . . , φ m ∈ L := { ( c 1 , . . . , c m ) : φ c 1 1 φ c 2 2 · · · φ c m � m . m = 1 } ⊆ This is a lattice. Let � [ x 1 , . . . , x m ] I ( L ) := � x c 1 1 x c 2 2 · · · x c m m − 1 : ( c 1 , . . . , c m ) ∈ L � � be the corresponding lattice ideal. (It is understood that “denominators are cleared”.) Let f 0 ( n ) = n and f 1 ( n ) = φ n 1 , f 2 ( n ) = φ n 2 , . . . , f m ( n ) = φ n m .
In General: Algebraic Relations of Exponentials � and Theorem: Let φ 1 , . . . , φ m ∈ L := { ( c 1 , . . . , c m ) : φ c 1 1 φ c 2 2 · · · φ c m � m . m = 1 } ⊆ This is a lattice. Let � [ x 1 , . . . , x m ] I ( L ) := � x c 1 1 x c 2 2 · · · x c m m − 1 : ( c 1 , . . . , c m ) ∈ L � � be the corresponding lattice ideal. (It is understood that “denominators are cleared”.) Let f 0 ( n ) = n and f 1 ( n ) = φ n 1 , f 2 ( n ) = φ n 2 , . . . , f m ( n ) = φ n m . Then
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